how Many grams of copper(11) acetate monohydrate will you ned to prepare 10 mL of a 0.01 M solution?
How many moles do you need? moles = M x L = ??
Then moles = grams/molar mass
Sole for grams.
To calculate the number of grams of copper(II) acetate monohydrate needed to make a 0.01 M solution in 10 mL, we need to use the formula:
moles = concentration × volume
First, let's convert the volume from milliliters (mL) to liters (L):
10 mL = 10/1000 L = 0.01 L
Next, let's rearrange the formula to solve for moles:
moles = concentration × volume
moles = 0.01 M × 0.01 L
moles = 0.0001 mol
Now, we need to find the molar mass of copper(II) acetate monohydrate. The formula for copper(II) acetate monohydrate is Cu(CH3COO)2 • H2O.
The molar mass for copper (Cu) is 63.55 g/mol.
The molar mass for carbon (C) is 12.01 g/mol.
The molar mass for hydrogen (H) is 1.008 g/mol.
The molar mass for oxygen (O) is 16.00 g/mol.
Since there are two moles of acetate (CH3COO) and one mole of water (H2O), we need to calculate the molar mass of copper(II) acetate monohydrate as follows:
(2 × molar mass of carbon) + (6 × molar mass of hydrogen) + (2 × molar mass of oxygen) + molar mass of copper + molar mass of water
(2 × 12.01 g/mol) + (6 × 1.008 g/mol) + (2 × 16.00 g/mol) + 63.55 g/mol + 18.015 g/mol = 199.63 g/mol
Now that we have the moles and molar mass, we can calculate the grams of copper(II) acetate monohydrate needed:
grams = moles × molar mass
grams = 0.0001 mol × 199.63 g/mol
grams ≈ 0.019963 g
Therefore, you would need approximately 0.019963 grams of copper(II) acetate monohydrate to prepare a 0.01 M solution in 10 mL.